3 Marks Question

Question 13 Marks

State the Lenz's law regarding electromagnetic induction. This law is based on the conservation law of which quantity?
A horizontal conducting wire of length 2 m oriented in east-west direction is falling down with a speed of 5 m/s perpendicular to the horizontal component of earth's magnetic field of $0.3 \times 10^{-4} T.$ Calculate the instantaneous value of induced emf across the ends of the wire.

Answer

Lenz's Law : According to this law, "The direction of induced emf (or say induced current) in a circuit is such that it always opposes the cause or say change due to which it is produced."
In the mathematical expression of Faraday's second law of electromagnetic induction negative sign is the representation of Lenz's law.
This law is based on the law of conservation of energy.
Solution of Numerical question :
Instantaneous value of induced emf across the ends of the wire:
$e=Bvl \sin\theta \Rightarrow e=B_{H}vl \sin 90^{\circ}=B_{H}vl$
$e=(0.3 \times 10^{-4} T) \times (5 m/s) \times (2 m)$
$=0.3 \times 10^{-3} volt = 0.3 mV$

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Question 23 Marks

Define coefficient of mutual induction i.e. mutual inductance.
When the current in a primary coil is changed from zero to 2.0 A within 300 ms, then an induced emf of a 0.80 volt is produced in secondary coil. Calculate coefficient of mutual inductance between the two coils.

Answer

Coefficient of mutual induction : It may be defined in following two ways:
(i) Coefficient of mutual induction between two coils is equal to the number of flux linkages in any one of the coils when the current in the second coil is one ampere.
(ii) The coefficient of mutual induction between two coils is equal to the numerical value of induced emf in one coil (secondary coil) which is generated due to unit rate to change of current i.e. 1 amp/sec in the other (secondary coil).
Solution of Numerical :
$|e_{2}|=M(\frac{\Delta I_{1}}{\Delta t}) \Rightarrow M=\frac{|e_{2}| \times \Delta t}{\Delta I_{1}}$
$=[\frac{(0.8) \times (0.3)}{(2-0)}]H$
$=0.12 H$

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Question 33 Marks

Write Faraday's laws of electromagnetic induction.

Answer

Faraday's Laws of electromagnetic induction :
(1) Whenever there is a change in the magnetic flux linked with a coil, emf is induced in the coil. This emf is called induced emf and it lasts so long as the change in the flux is taking place. When the coil circuit is closed an electric current flows through the circuit corresponding to induced emf and is known as induced current. This phenomenon is called electromagnetic induction.
(2) The induced emf is equal to the negative rate of change of magnetic flux.
If $\Delta\Phi$ be the change in magnetic flux linked with a circuit in time interval $\Delta t$ then the induced emf in the circuit according to second law is given by:
$e=-(\frac{\Delta\Phi}{\Delta t})$ ...(1)
Under the condition: lim $\Delta t \rightarrow 0$ equation (1) in its differential form can be expressed as follows:
$e=-(\frac{d\Phi}{dt})$ ...(2)
The (-ve) sign indicates that the induced emf opposes the change in magnetic flux.

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Question 43 Marks

Obtain the expression for self inductance of an ideal solenoid.

Answer

Magnetic field at any point on its axis inside the long solenoid is uniform and is given by :
$B=\frac{\mu_{0}NI}{l}$
where $l=$ Length of the solenoid; $N=$ Number of turns in the total length of the solenoid and I is the current through the solenoid.
Therefore the magnetic flux linked with each turn of the solenoid is given by :
$\Phi=B \times A$
where $A=\pi r^{2}=$ Arc of each turn in which $r$ is the radius of the solenoid.
$\therefore$ $\Phi=\frac{\mu_{0}NI}{l} \times \pi r^{2} \Rightarrow \Phi=\frac{\mu_{0}\pi Nr^{2}I}{l}$
But total magnetic flux linked with the solenoid is $N\Phi$ which is given by :
$N\Phi=LI$
$\Rightarrow N \times (\frac{\mu_{0}\pi Nr^{2}I}{l})=LI$
⇒ $L=\frac{\mu_{0}\pi N^{2}r^{2}}{l}\quad $ ...(1)
This is the required expression and can also be written as fallows :
$L=\mu_{0}\pi\frac{N^{2}}{l^{2}}r^{2}.l = \mu_{0}(\pi r^{2})(N/l)^{2}.l$
But $\pi r^{2}=A$ and $(N/l)=n=$ number of terms per unit length
$\therefore$ $L=\mu_{0}An^{2}l\quad$ ...(2)

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